limiti notevoli delle successioni e forme indeterminateInformatica Sapienza

Se la successione $a_{n} \to \infty, A \in R$ $lim_{n \to + \infty} (1 + \frac{A}{a _{n}})^{a_{n}} = e^{A}$

$\frac{0}{0}$

$lim_{n \to + \infty} \frac{a _{n}}{b _{n}}, b_{n} \to 0, a_{n} \to L \neq = 0$

Se $0 \leq b_{n} \to 0 ⟹ \frac{a _{n}}{b _{n}} \to - \infty$ Se $0 \geq b_{n} \to 0 ⟹ \frac{a _{n}}{b _{n}} \to + \infty$

Se $0 \leq b_{n} \to 0 ⟹ \frac{a _{n}}{b _{n}} \to + \infty$ Se $0 \geq b_{n} \to 0 ⟹ \frac{a _{n}}{b _{n}} \to - \infty$

Sia $a_{n} \to 0$

$lim_{n \to + \infty} \frac{e ^{a_{n}} - 1}{a _{n}} = 1$	$lim_{n \to + \infty} \frac{l o g ( 1 + a _{n} )}{a _{n}} = 1$
$lim_{n \to + \infty} \frac{s i n ( a _{n} )}{a _{n}} = 1$	$lim_{n \to + \infty} \frac{t a n ( a _{n} )}{a _{n}} = 1$
$lim_{n \to + \infty} \frac{1 - c o s ( a _{n} )}{a _{n}} = 0$	$lim_{n \to + \infty} \frac{1 - c o s ( a _{n} )}{a _{n}^{2}} = \frac{1}{2}$
$lim_{n \to + \infty} \frac{a r c t a n ( a _{n} )}{a _{n}} = 1$	$lim_{n \to + \infty} \frac{a r c s i n ( a _{n} )}{a _{n}} = 1$
Se $a_{n} \to \infty, A \in R$
$lim_{n \to + \infty} (1 + \frac{A}{a _{n}})^{a_{n}} = e^{A}$